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Home » The     and     terms of a geometric series are     and sum     , respectively. Find the sum of the series up to     terms.
The

    \[{{2}^{nd}}\]

and

    \[{{5}^{th}}\]

terms of a geometric series are

    \[-{\scriptscriptstyle 1\!/\!{ }_2}\]

and sum

    \[\mathbf{1}/\mathbf{16}\]

, respectively. Find the sum of the series up to

    \[8\]

terms.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
The

    \[{{2}^{nd}}\]

and

    \[{{5}^{th}}\]

terms of a geometric series are

    \[-{\scriptscriptstyle 1\!/\!{ }_2}\]

and sum

    \[\mathbf{1}/\mathbf{16}\]

, respectively. Find the sum of the series up to

    \[8\]

terms.

From the question it is given that,

    \[\begin{array}{*{35}{l}} {{a}_{2}}~=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2} \\ {{a}_{5}}~=\text{ }1/16 \\ \end{array}\]

We know that,

    \[{{a}_{2}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\]

    \[=\text{ }a{{r}^{2\text{ }\text{ }1}}\]

    \[{{a}_{2}}~=\text{ }ar\text{ }=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2}\]

… [equation (i)]

    \[\begin{array}{*{35}{l}} {{a}_{5}}~=\text{ }a{{r}^{5\text{ }\text{ }1}} \\ =\text{ }a{{r}^{4}} \\ \end{array}\]

    \[{{a}_{5}}~=\text{ }a{{r}^{4}}~=\text{ }1/16\]

… [equation (ii)]

Now, dividing equation (ii) by (i) we get,

    \[\begin{array}{*{35}{l}} {{r}^{3}}~=\text{ }1/16\text{ }\div \text{ }\left( -{\scriptscriptstyle 1\!/\!{ }_2} \right) \\ =\text{ }\left( 1/16 \right)\text{ }\times \text{ }\left( -2 \right) \\ =\text{ }-1/8 \\ {{r}^{3}}~=\text{ }{{\left( -1/2 \right)}^{3}} \\ \end{array}\]

So,

    \[\begin{array}{*{35}{l}} ~r\text{ }=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2} \\ ar\text{ }=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2} \\ a\text{ }\times \text{ }\left( -{\scriptscriptstyle 1\!/\!{ }_2} \right)\text{ }=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2} \\ a\text{ }=\text{ }\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\times \text{ }\left( -2/1 \right) \\ a\text{ }=\text{ }1 \\ \end{array}\]

Therefore,

    \[a\text{ }=\text{ }1\text{ }and\text{ }r\text{ }=\text{ }-{\scriptscriptstyle 1\!/\!{ }_2}\]

Then,

    \[\begin{array}{*{35}{l}}</strong> <strong> {{S}_{8}}~=\text{ }a(1\text{ }\text{ }{{r}^{n}})/\left( 1\text{ }\text{ }r \right) \\</strong> <strong> =\text{ }1[1\text{ }\text{ }{{\left( -1/2 \right)}^{8}}]/\left( 1\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_2} \right) \\</strong> <strong> =\text{ }\left[ 1\text{ }\text{ }\left( 1/256 \right) \right]/\left( 3/2 \right) \\</strong> <strong> =\text{ }\left( 255/256 \right)\text{ }\times \text{ }\left( 2/3 \right) \\</strong> <strong> =\text{ }\left( 510/768 \right) \\</strong> <strong> =\text{ }85/128 \\</strong> <strong>\end{array}\]

i

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